HW 3 - Analysis 1 - MATH 3380.tex

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\begin{document}

Assignment Set: {6, 7, 15, 17, 19, 21} from pages 141 - 142

\bsc{6)}

Find the closure of each set:

\balist
\item \{ $\frac{1}{n}$ : n $\in \bn$ \} \

	Answer: $\es$
\item $\bn$ \
	
	Answer: $\bn$
\item $\bq$ \
	
	Answer: $\br$
\item $\bigcap_{n=1}^\infty$ $(0, \frac{1}{n})$ \
	
	Answer: $\es$
\item \{ x : $|x - 5| \leq \frac{1}{2}$\} \

	$[4.5, 5.5]$ \
	
	Answer: $[4.5, 5.5]$
\item \{ x : $x^2 > 0$\} \
	
	$(0, \infty)$ \
	
	Answer: $[0, \infty)$ \
\elist

\bsc{7)} Let S, T $\sbs \br$. Find a counterexample of each of the following:
\balist
\item If P is the set of all isolated points of S, then P is a closed set. \

	Answer: Let S = $\bn$
\item Every open set contains at least two points. \

	Answer: $\es$
\item If S is closed, then cl(int S) $=$ S. \

	Answer: Let S $= \bq$
\item If S is open, then int (cl S) $=$ S. \

	Answer: Let S $= (-1, 0) \cup (0, 1)$
\item bd (cl S) $=$ bd S \

	Answer: Let S $= (-1, 0) \cup (0, 1)$
\item bd (bd S) $=$ bd S \

	Answer: Let S = $\bq$. Then bd S is $\br$, and bd (bd S) $= \es \neq \br$.
\item bd (S $\cup$ T) $=$ (bd S) $\cup$ (bd T) \

	Answer: Let S $= \br$, T $= (0, 1)$. bd (S $\cup$ T) = $\es$, but bd S $\cup$ bd T $= \es \cup \{0, 1\}$
\item bd (S $\cap$ T) $=$ (bd S) $\cap$ (bd T) \

	Answer: Let S $= (0, 1)$, T $= (1, 2)$. bd (S $\cap$ T) $= \es$, but bd S $\cap$ bd T $=$ 1.
\elist

\bsc{15)} Prove: If x is an accumulation point of the set S, then every neighborhood of x contains infinitely many points of S.\

\bgpf
Suppose that $\exists$ a deleted neighborhood of x, called N, that contains n points x$_1$, x$_2$, ... x$_n$ of S where n is a finite amount and x$_1$ $\leq$ x$_2$, $\leq$  ... x$_n$ \

x is an accumulation point on S if $\fa \ep > 0$, $\dnbho{x}{\epsilon}{S	}$. \

N is a deleted neighborhood of S if $\fa$ x $\in \{y \in \br : 0 < |y - x| < \ep\}$, x $\in$ N.\

Let $\hat\epsilon = \ep + \ep$, and x$_0$ = x$_1$ - $\hat\epsilon$. \

By definition, x$_0 \in$ N, since N is a neighborhood $\fa \ep > 0$.\

However, N only has n elements. A contradiction.\

So, N can't be a deleted neighborhood since it has a finite number of elements, which means x can't be an accumulation point.

\epf
 

\bsc{17)} Prove: S\pr is a closed set.

\bgpf

By definition, $\fa x \in S\pr$, \ep $> 0$, $\dnbho{x}{\epsilon}{S}$

Notice that if S\pr is empty or S\pr is \br, then S\pr is a closed set and we are done. \


\lt{\bnt{\br}{S'} \sbs \br, x \mem \bnt{\br}{S'}}



\wts{\bnt{\br}{S\pr} is open.}

\bnt{\br}{S\pr} is open iff \bnt{\br}{S\pr} \eql int (\bnt{\br}{S\pr})

int \bnt{\br}{S\pr} \eql \{s: N(s, \ep) \sbs \bnt{\br}{S\pr}\}



\epf

\bsc{19)} Suppose S is a nonempty bounded set and let m $=$ sup S. Prove or give a counter example: m is a boundary point of S.

\bgpf

By definition,

s $\leq$ m, $\fa$ s $\in$ S, and,

$\fa$ $\ep > 0$, $\exists$ s' $\in$ S st m $-$ $\ep <$ s'

By the second part of the definition of the supremum of S, $\nbho{m}{\epsilon}{S}$.

Notice also that, by the first part of the definition of the supremum of S, (m $+$ $\ep) \not\in$ S. This means that $\nbho{m}{\epsilon}{\bnt{\br}{S}}$.

By definition, m is a boundary point.



\epf

\pagebreak

\bsc{21)} Let A be a nonempty open subset of $\br$ and let Q $\sbs \bq$. Prove: A $\cap$ Q $\neq \es$.

\bgpf
Notice that Q $\sbs \bq \sbs \br$. \

Since A is nonempty, $\exists$ at least one element a $\in \br$. \

Since A is nonempty and open, a $+$ $\ep \in$ A. \

If a $\in \bq$, then result. \

If a $+ \ep \in \bq$, then result. \

If a $\not\in \bq$ and (a $+$ $\ep) \not\in \bq$, then: \

Let x $=$ a, y $=$ a $+$ $\ep$, z $=$ y $-$ x. \

By Archimedes' axiom, $\exists$ n st n $> \frac{1}{z	}$ \

nz $>$ 1

ny $-$ nx $>$ 1

Since the difference between ny and nx is bigger than 1, \

$\exists m \in \bz$ st nx $<$ m $<$ ny.

See that since  x $< \frac{m}{n	} < y$, $\frac{m}{n}$ is a rational number, and $\frac{m}{n} \in$ A. \

Hence, result. 
\epf

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